# Functional Equation with a hint of differentiability

Okay so I got this question in a rehearsal test today , and got completely stumped.

Let $f$ be a differentiable function satisfying the functional rule $f(xy)=f(x)+f(y)+\frac{x+y-1}{xy}$, $\forall$ $x,y>0$ and $f'(1)=2$.The question asked some more questions regarding $f(x)$,but I am unable to find $f(x)$.

My try:I tried to treat $y$ as a constant and then differentiated both sides of the equation,making use of the fact that $f'(1)=2$ ; but that got me nowhere.Someone kindly help.

• try substituting y=1/x and the differentiate – Savitar May 10 '17 at 15:10
• How does that help ? – Zlatan May 10 '17 at 15:41

$$yf'(xy) = f'(x) -\frac{1}{x^2} + \frac{1}{x^2y},\tag{1}$$

for all $x,y > 0$. Choosing $x = 1$ in $(1)$, we obtain

$$yf'(y) = 2 - 1 + \frac{1}{y},$$

or

$$f'(y) = \frac{1}{y} + \frac{1}{y^2}.\tag{2}$$

From that, we can easily determine $f$ up to addition of a constant, and we can determine the constant from the rule e.g. by setting $x = y = 1$ there.

An alternative way to find the function is to rewrite the functional equation as

$$f(xy) + \frac{1}{xy} = \biggl(f(x) + \frac{1}{x}\biggr) + \biggl(f(y) + \frac{1}{y}\biggr),$$

and if one knows the continuous functions satisfying the functional equation $g(xy) = g(x) + g(y)$, $f$ is easily determined.

• Sorry , Thanks for your effort into this , but like a noob I am ,Im not getting anything.How do i get $f(x)$ from this ? And how do I tackle similar problems?Isnt there a thumb rule? – Zlatan May 10 '17 at 15:50
• We have a nice form for the derivative of $f$. One can read the primitives of $f'$ off that if one knows the derivatives of the most famous functions. If $g'(y) = y^{\alpha}$, what is $g(y)$? – Daniel Fischer May 10 '17 at 16:01

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{f}}$ is differentiable. $\ds{\mrm{f}\pars{xy} = \,\mrm{f}\pars{x} + \,\mrm{f}\pars{y} + {1 \over y} + {1 \over x} - {1 \over xy}\,,\quad x, y \in \mathbb{R}_{\ >\ 0}\quad\mbox{and}\quad\mrm{f}'\pars{1} = 2}$.

$$\bbx{\quad\mbox{Note that}\quad\mrm{f}\pars{1} = -1\quad \pars{~\mbox{from the 'original functional equation'}~}\quad}$$

• Derive both members, of the above functional equation, respect of $\ds{x}$: $$\mrm{f}'\pars{xy}y = \,\mrm{f}'\pars{x} - {1 \over x^{2}} + {1 \over x^{2}y} \implies \mrm{f}'\pars{xy}xy = \,\mrm{f}'\pars{x}x - {1 \over x} + {1 \over xy} \label{1}\tag{1}$$
• Derive both members, of the above functional equation, respect of $\ds{y}$: $$\mrm{f}'\pars{xy}x = \,\mrm{f}'\pars{y} - {1 \over y^{2}} + {1 \over xy^{2}} \implies \mrm{f}'\pars{xy}xy = \,\mrm{f}'\pars{y}y - {1 \over y} + {1 \over xy} \label{2}\tag{2}$$ \eqref{1} and \eqref{2} lead to: \begin{align} &\,\mrm{f}'\pars{x}x - {1 \over x} = \,\mrm{f}'\pars{y}y - {1 \over y} = \alpha\,, \qquad\pars{~\alpha:\ \mbox{a}\ x\ \mbox{and}\ y\ \mbox{independent constant}~} \end{align}

$\ds{\alpha = \,\mrm{f}'\pars{1} \times 1 - {1 \over 1} = 1}$.

Now, you are left with $$\,\mrm{f}'\pars{x} = {1 \over x} + {1 \over x^{2}}\,,\quad\mrm{f}\pars{1} = -1 \implies \bbox[#ffe,10px,border:1px dotted navy]{\ds{\mrm{f}\pars{x} = \ln\pars{x} - {1 \over x}}}$$